Hundred Practice/2016 computer Science summer camp on-Machine exam: G

Title Source: http://dsalgo.openjudge.cn/binarytree/8/ 8: Rebuilding the binary tree

Total time limit: 1000ms memory limit: 65536kB

Description

Given a binary tree's pre-sequence traversal and the result of the middle sequence traversal, its post-order traversal is obtained.

input

The input may have more than one group, ending with EOF.
Each set of inputs contains two strings, respectively, for the tree's pre-order traversal and the middle sequence traversal. Each string contains only uppercase letters and is distinct from each other.

Output

For each set of inputs, a row is used to output its post-order traversal results.

Sample Input

DBACEGF ABCDEFG

Bcad Cbad

Sample Output

Acbfged

Cdab

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Thinking of solving problems

Recursive, first locating the root node and then slicing the Saozi right subtree

Note the application of the String.substr () method

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Code

#include <iostream>
#include <string>
using namespace std;

String Postorder (string mid, String Pre)			//input preorder string and inorder string, output Postorder string
{
	if (pre.length () = = 1) return pre;					If preorder has only one character then preorder = Inorder = Postorder
	Else if (pre.length () ==0) return "";				Returns an empty string if preorder is empty/
	/This avoids the existence of a left/right subtree discussion
	else
	{
		size_t root = Mid.find (pre.at (0));			The root node of the tree
		return Postorder (Mid.substr (0,root), Pre.substr (1,root))			//First left subtree
			+ postorder (MID.SUBSTR ( root+1), Pre.substr (root+1)) + pre.at (0);//Right Sub-tree, last root node
		//Note: Str.substr (str.size ()) returns an empty string without throwing an exception.
	}
}

int main ()
{
	string Pre, Mid, post;//s1 to preorder traversal  S2 to inorder traversal while 
	(cin > > Pre >> mid)
	{
		Post=postorder (Mid, pre);
		cout << post << Endl;
		
	}
	return 0;
}